Prove $\mid$x-1$\mid$+$\mid$x+5$\mid$ $\ge$ 6 for all real numbers

Question

I need to prove it with proof by cases. I graphed it out and it seems that I need to put into 2 cases which is when x$>$1 and x$>$-5. But this where my problem is, how do plug that back into the statement? Would it just be calculating (x-1)+(x+5)?

Answer 1

We have, the real function of the real variable $f(x)=\sum_{i=1}^n |x-\alpha_i|,$ for which $\alpha_1\geq \dots \geq \alpha_n$ are real numbers. Then,

$ 1)$ If $n$ is odd, then $\min_{x\in\mathbb{R}} f(x)=(\alpha_1+\cdots+\alpha_{\frac{n-1}{2}})-(\alpha_{\frac{n+3}{2}}+\cdots+\alpha_n),$ and minimum is reached when $x=\alpha_{\frac{n+1}{2}}$

$2)$ If $n$ is even, then $\min_{x\in\mathbb{R}} f(x)=(\alpha_1+\cdots+\alpha_{\frac{n}{2}})-(\alpha_{\frac{n}{2}+1}+\cdots+\alpha_n),$ and this minimum is reached for every $x\in[\alpha_{\frac{n}{2}}, \alpha_{\frac{n}{2}+1}]$ if $\alpha_{\frac{n}{2}} \neq \alpha_{\frac{n}{2}+1} $

and attains minimum only for $x=\alpha_{\frac{n}{2}}$ if $\alpha_{\frac{n}{2}} = \alpha_{\frac{n}{2}+1}. $ The above can be proved using calculus.

The given problem is particular case of this. Also can be solved by taking the function as follows.

$ f(x)= \begin{cases} x-1+x+5,& \text{if } x\geq 1\\ 6, &\text{if } x-5 \leq x \leq 1\\ 1-x+5-x,&\text{if } x \leq -5 \end{cases}$ and then calculate critical points and maxima and minimum of this function.

Answer 2

I think there is a typo, and it should read: " $|x+1|+|x-5| \ge 6$". And for this we have: $|x+1| + |x-5| = |-x - 1|+|x-5| \ge |-x-1+x-5| = |-6| = 6$ , by the $\triangle$ inequality.

Answer 3

$y:=x-1$, $y$ real.

Need to show :

$|y|+|y+6| \ge 6.$

$|6| = |(6+y)-y| \le |6+y| +|y|$.

(Triangle inequality).

Answer 4

We can rewrite this as

$$ |x-1| + |x-(-5)| \geq 6 $$

and then it becomes apparent that geometrically, we are looking at the sum of two distances on the number line:

• the distance between $x$ and $1$, and
• the distance between $x$ and $-5$

If $x$ is between $-5$ and $1$, then this is just the distance between $-5$ and $1$, and if it's less than $-5$ or greater than $1$, it should be greater than that distance. Since the distance between $-5$ and $1$ is already $6$, the sum must be greater than or equal to $6$.